Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Noether Theorem

  1. Dec 17, 2015 #1
  2. jcsd
  3. Dec 17, 2015 #2

    samalkhaiat

    User Avatar
    Science Advisor

    Which step you don't understand? Equation (7) follows from Tylor expanding the Lagrangian:
    [tex]L(x + \epsilon , y+ \eta) - L(x,y) = L(x,y) + \frac{\partial L}{\partial x} \epsilon + \frac{\partial L}{\partial y} \eta - L(x,y)[/tex]
    The left hand side is the variation in L, i.e., [itex]\delta L[/itex]. Now take [itex]x = q[/itex], [itex]y = \dot{q}[/itex], [itex]\epsilon = \gamma[/itex] and [itex]\eta = \dot{\gamma}[/itex], you get
    [tex]\delta L = \frac{\partial L}{\partial q} \gamma + \frac{\partial L}{\partial \dot{q}} \dot{\gamma} .[/tex]
    If the transformation [itex]q^{'} = q + \gamma[/itex] is a symmetry, then [itex]\delta L = 0[/itex]. So, you have
    [tex]\frac{\partial L}{\partial q} \gamma + \frac{\partial L}{\partial \dot{q}} \dot{\gamma} = 0.[/tex]
    Now, use the Lagrange equation
    [tex]\frac{\partial L}{\partial q} = \frac{d}{dt}( \frac{\partial L}{\partial \dot{q}} ) ,[/tex]
    in the first term and combine the two terms
    [tex]\frac{d}{dt}( \frac{\partial L}{\partial \dot{q}} ) \gamma + (\frac{\partial L}{\partial \dot{q}} ) \frac{d}{dt}\gamma = \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}} \gamma \right) = 0 .[/tex]
     
  4. Dec 18, 2015 #3
    Thanks! But can you show me taylor expansion of the Lagrangian? I am kind of new to this.
     
  5. Dec 18, 2015 #4

    samalkhaiat

    User Avatar
    Science Advisor

    This is calculus problem! You should be able to Taylor expand function of two variables:
    [tex]L(q + \epsilon \gamma , \dot{q} + \epsilon \dot{\gamma}) = L(q,\dot{q}) + \epsilon \gamma \frac{\partial L}{\partial q} + \epsilon \dot{\gamma} \frac{\partial L}{\partial \dot{q}} + \frac{1}{2} \epsilon^{2} \gamma^{2} \frac{\partial^{2}L}{\partial q^{2}} + \cdots[/tex]
    Infinitesimal transformations means that [itex]\epsilon \ll 1[/itex], so you can take [itex]\epsilon^{2} = \epsilon^{3} = \cdots \approx 0[/itex] and keep only the linear terms in [itex]\epsilon[/itex]:
    [tex]L(q + \epsilon \gamma , \dot{q} + \epsilon \dot{\gamma}) = L(q,\dot{q}) + \epsilon \gamma \frac{\partial L}{\partial q} + \epsilon \dot{\gamma} \frac{\partial L}{\partial \dot{q}}[/tex]
     
  6. Dec 19, 2015 #5
    Thanks! But how do I get to equation 32?
     
  7. Dec 19, 2015 #6

    samalkhaiat

    User Avatar
    Science Advisor

    Don't you know how to take the total time derivative of [itex]L(x(t),y(t);t)[/itex]? What is your formal Education level?
    [tex]\frac{dL}{dt} = \frac{\partial L}{\partial t} + \frac{\partial L}{\partial x} \frac{dx}{dt} + \frac{\partial L}{\partial y} \frac{dy}{dt} [/tex]
    Now take [itex]x=q, \ y = \frac{dx}{dt} = \frac{dq}{dt} \equiv \dot{q}[/itex], so [itex]\frac{dy}{dt} = \ddot{q}[/itex].
     
  8. Dec 19, 2015 #7
    Sorry sometimes when I do too much my brain gets fuzzy and I tend to forget all my stuff. I just have one last question, as to how the author just come up with this equation
    upload_2015-12-20_8-49-29.png

    Thanks again for your help! Besides i am a A'level student.
     
  9. Dec 20, 2015 #8
    Sorry. Forget my previous post which was a stupid question. My last question is that in this case, is the degree of freedom referring to the type of axis(e.g. x, y or z) or the position along one axis for the euclidean translational symmetry on this paper?
     
  10. Dec 20, 2015 #9

    samalkhaiat

    User Avatar
    Science Advisor

    It may or may not be. As you are still an A-level student, I would suggest you wait till you are mathematically more able. Or, if you like, you can start by reading about the meaning of generalized coordinates in analytical Mechanics.
    Good Luck
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Noether Theorem
  1. Noether's Theorem (Replies: 36)

  2. Noether's Theorem (Replies: 3)

Loading...